Question

The elements A and Z combine to produce two different compounds: \(\mathrm{A}_{2} \mathrm{Z}_{3}\) and \(\mathrm{AZ}_{2}\). If 0.15 mol of \(\mathrm{A}_{2} \mathrm{Z}_{3}\) has a mass of \(15.9 \mathrm{g}\) and 0.15 mole of \(\mathrm{AZ}_{2}\) has a mass of \(9.3 \mathrm{g}\) what are the atomic masses of \(\mathrm{A}\) and \(\mathrm{Z} ?\)

Short answer

The atomic masses are 26 g/mol for A and 18 g/mol for Z.

Step-by-step solution

Determine Molar Mass of Compounds

First, calculate the molar mass of each compound using the mass and moles given. The molar mass is calculated as mass divided by the number of moles. For \(\mathrm{A}_{2} \mathrm{Z}_{3}\), the molar mass is \(\frac{15.9 \text{ g}}{0.15 \text{ mol}} = 106 \text{ g/mol}\). For \(\mathrm{AZ}_{2}\), it is \(\frac{9.3 \text{ g}}{0.15 \text{ mol}} = 62 \text{ g/mol}\).

Model Equations for Molar Masses

Using the formulas for each compound, set up equations for their molecular masses: \(2M_A + 3M_Z = 106\) for \(\mathrm{A}_{2} \mathrm{Z}_{3}\) and \(M_A + 2M_Z = 62\) for \(\mathrm{AZ}_{2}\). Here, \(M_A\) and \(M_Z\) represent the atomic masses of elements A and Z.

Solve the System of Equations

We now solve the system of equations: \(2M_A + 3M_Z = 106\) and \(M_A + 2M_Z = 62\). Multiply the second equation by 2: \(2M_A + 4M_Z = 124\). Subtract the first equation from this new equation: \((2M_A + 4M_Z) - (2M_A + 3M_Z) = 124 - 106\), which simplifies to \(M_Z = 18\). Substitute \(M_Z = 18\) into the second original equation: \(M_A + 2(18) = 62\), so \(M_A = 62 - 36 = 26\).

Verify the Calculated Atomic Masses

Using \(M_A = 26\) and \(M_Z = 18\), compute the molar masses from the original equations: \(2(26) + 3(18) = 106\) and \(26 + 2(18) = 62\). Both calculations agree with the given compound molar masses, confirming that \(M_A\) and \(M_Z\) are correct.

Key concepts

Molar Mass Determination

Molar mass is a fundamental concept in chemistry that refers to the mass of a compound divided by the number of moles. It's a way to express the amount of substance contained in one mole of a compound, and it's crucial for determining the relationship between mass and number of particles. In the original exercise, the compounds \(\mathrm{A}_{2}\mathrm{Z}_{3}\) and \(\mathrm{AZ}_{2}\) have their molar masses determined by dividing the provided mass by the number of moles (0.15 mol). This gives:

• For \(\mathrm{A}_{2}\mathrm{Z}_{3}\), the molar mass is \(\frac{15.9 \text{ g}}{0.15 \text{ mol}} = 106 \text{ g/mol}\).
• For \(\mathrm{AZ}_{2}\), it is \(\frac{9.3 \text{ g}}{0.15 \text{ mol}} = 62 \text{ g/mol}\).

These molar masses serve as a crucial stepping stone for deriving the atomic masses of the individual elements A and Z through further analysis of the chemical compounds.

Chemical Equation Solving

Solving chemical equations is an essential skill when working with reactions and compound formulations. Once the molar masses of \(\mathrm{A}_{2}\mathrm{Z}_{3}\) and \(\mathrm{AZ}_{2}\) were determined, it was necessary to form equations that express these masses in terms of the atomic masses of A and Z. By examining the compound formulas:

• For \(\mathrm{A}_{2}\mathrm{Z}_{3}\), we write the equation as \(2M_A + 3M_Z = 106\).
• For \(\mathrm{AZ}_{2}\), we have \(M_A + 2M_Z = 62\).

Here, \(M_A\) and \(M_Z\) denote the atomic masses of A and Z. These equations form a system of linear equations, which allows us to solve for the unknowns, \(M_A\) and \(M_Z\), using algebraic methods.

Compound Analysis

Compound analysis focuses on understanding the composition and properties of compounds, allowing us to deduce the atomic structures within them. During the exercise, solving the system of equations involves algebraic manipulation to uncover the atomic masses of elements A and Z. The steps involve:

• Multiplying the equation for \(\mathrm{AZ}_{2}\) by 2 gives \(2M_A + 4M_Z = 124\).
• Subtracting it from the \(\mathrm{A}_{2}\mathrm{Z}_{3}\) equation leads to \(M_Z = 18\).
• Substituting \(M_Z = 18\) in the equation \(M_A + 2M_Z = 62\) solves to \(M_A = 26\).

These values are then verified by ensuring they satisfy both original equations, confirming that the analysis and determination were correct. This methodical approach to compound analysis is crucial for understanding chemical compositions and reactions accurately.